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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$.
For the first source: $\frac{hc

Vedclass · Accepted Answer

$1.72 	imes 10^{-7} \ m, 1.20 \ eV$

Vedclass · Answer

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$.
For the first source: $\frac{hc}{\lambda} = \phi + 6.0 \ eV$ ... $(i)$
For the second source: $\frac{hc}{4\lambda} = \phi + 0.6 \ eV$ ... (ii)
Multiply equation (ii) by $4$: $\frac{hc}{\lambda} = 4\phi + 2.4 \ eV$ ... (iii)
Equating $(i)$ and (iii): $\phi + 6.0 = 4\phi + 2.4 \Rightarrow 3\phi = 3.6 \Rightarrow \phi = 1.2 \ eV$.
Substitute $\phi$ into $(i)$: $\frac{hc}{\lambda} = 1.2 + 6.0 = 7.2 \ eV$.
Given $hc = 1.24 	imes 10^{-6} \ J \ m = \frac{1.24 	imes 10^{-6}}{1.6 	imes 10^{-19}} \ eV \ m = 7.75 	imes 10^{12} \ eV \ m$ (approx) or using $hc = 1240 \ eV \ nm = 1.24 	imes 10^{-6} \ eV \ m$.
$\lambda = \frac{1.24 	imes 10^{-6}}{7.2} \approx 1.72 	imes 10^{-7} \ m$.

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